A 2 lb box slides on the smooth curved ramp. If the box has a velocity of 20 ft/s at A, determine the velocity of the box when it is located at B and also find the normal force acting on the ramp at B. Assume the radius of curvature of the path at C is still 5 ft.

1) 8.8 m & 0.96 lb
2) 3.5 m & 0.34 lb
3) 4.5 m & 0.65 lb
4) 6.5 m & 0.50 lb

Summing up the forces in all the directions we have:
T1 + S U1-2 = T2
1/2 (2/32.2) (20)2 - 2 (5) = 1/2 (2/32.2) (VB)2
VB= 8.8 ft/s

So the velocity at B is VB= 8.8 ft/s

At point B the summing up all the forces acting :
S Fn = man
NB = (2/32.2) (8.8)2/5
NB = 0.962 lb

So the velocity of the crate at B is VB= 8.8 ft/s and
the normal force acting at B is NB = 0.96 lb